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Problem 36

Solving a Trigonometric

Equation In Exercises $35-42,$ solve the equation for $\theta,$ where $0 \leq \theta \leq 2 \pi .$

$$\tan ^{2} \theta=3$$

Answer

$\theta=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}$

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## Discussion

## Video Transcript

this problem asks us to solve the following equation for theta, where theta is greater than zero Good, then equal to zero and less than equal to two. Now to begin, our equation is changing. Squared data equals three. And so what we will do is we will take the square root of both sides just so we can get one step closer to isolating data. Now we need to remember to do plus or minus, and we take the square root of three. Now, this is going to simplify two tangent data because, plus or minus square root of three. Now, what we can do next is come over to our unit circle and think about all our radiant measures where tangent data equals plus or minus the square root of three. And if you're familiar with Eunice circle the value of tangent vehicles plus or minus three occurs in multiples of high thirds, excluding pi and two pine. All right, so I've identified all the locations, so a pie thirds tangent data equals the square root of three. Here it was negative. Scurry three here in equal square three and here it equals negative square three. So when we find our answer. We can say that data equals hi thirds to buy thirds for by thirds and five by thirds. That's your answer.

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$$

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Solve the given trigonometric equation exactly on $0 \leq \theta<2 \pi$.

$$\tan ^{2} \theta-\sqrt{3} \tan \theta=0$$